MathDB

34

Part of 1965 AMC 12/AHSME

Problems(1)

Minimum Value

Source:

1/12/2009
For x≥0 x \ge 0 the smallest value of \frac {4x^2 \plus{} 8x \plus{} 13}{6(1 \plus{} x)} is: <spanclass=′latex−bold′>(A)</span> 1<spanclass=′latex−bold′>(B)</span> 2<spanclass=′latex−bold′>(C)</span> 2512<spanclass=′latex−bold′>(D)</span> 136<spanclass=′latex−bold′>(E)</span> 345 <span class='latex-bold'>(A)</span>\ 1 \qquad <span class='latex-bold'>(B)</span>\ 2 \qquad <span class='latex-bold'>(C)</span>\ \frac {25}{12} \qquad <span class='latex-bold'>(D)</span>\ \frac {13}{6} \qquad <span class='latex-bold'>(E)</span>\ \frac {34}{5}
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